If we have an order book and we assume that we know there is only one market maker, how can we determine exactly the spread of the market maker? What if there are more than one market makers?

  • 1
    $\begingroup$ You know how to determine the spread, it is Ask minus Bid. And if there is only one market maker that must be the "spread of the marketmaker". And if there are many (X, Y, Z, ...) then you can find the Ask of markemaker X minus the Bid of Marketmaker X. $\endgroup$
    – nbbo2
    Jul 15, 2020 at 18:35

1 Answer 1


Model Building

I will answer this question from a statistical perspective since that has a definitive answer (under stated assumptions)

assumption 1) Suppose that each market-maker applies the same, unknown, spread, $X$ to quote a specific product.

assumption 2) Suppose that the uncertainty in the mid-market price of each marker is determined by a normal distribution about a mean with a variance, $\mathcal{N}(\mu, \sigma^2)$

Then the quoted bid-ask spread is determined by:

  • Bid: Maximum mid - $\frac{X}{2}$
  • Ask: Minimum mid + $\frac{X}{2}$

The expectation of this can be assessed and depends upon the number, $n$, of market makers. This answer: https://math.stackexchange.com/questions/473229/expected-value-for-maximum-of-n-normal-random-variable is helpful. We can then derive the result:

  • $n = 1$: the tradeable bid-ask is $X$
  • $n = 2$: the expected tradeable bid-ask is: $$\underbrace{\left ( \mu - \sigma \pi^{-\frac{1}{2}} + \frac{X}{2} \right )}_{ask} - \underbrace{\left ( \mu + \sigma \pi^{-\frac{1}{2}} - \frac{X}{2} \right )}_{bid} = X - 2\sigma\pi^{-\frac{1}{2}}$$
  • $n = 3$: the expected tradeable bid-ask is: $X - 3\sigma\pi^{-\frac{1}{2}}$

The approximation given in the above link suggests that, in general, the expected tradeable bid-ask is given by:

$$ X - \sqrt{8 log(n)} \sigma$$


I actually think this is quite interesting. Take for example a market I know well - interest rate swaps - where there might be say 10 market makers and the tradeable bid-ask is typically 0.1 bps. If the uncertainty standard deviation is considered the same value then using the above approximation that each market maker would have an individual bid-ask of 0.38bps, which might be a bit wide but not unreasonably.

  • $\begingroup$ This is a very nice way to look at the problem. There is a slight concern in that delays may effectively increase $\sigma$, but that too can be determined. This is especially useful in markets where there are large numbers of market makers. I do wonder if this would help explain the prevalence of crossed markets near the close -- since $\sigma$ increases as well as $n$, perhaps to the point that $\sigma\sqrt{8\log(n)}>X$. Only caveat would be that in markets where delays do not dominate, crossed markets are very rare. $\endgroup$
    – kurtosis
    Aug 1, 2020 at 4:56
  • $\begingroup$ @kurtosis I missed this comment originally but I agree with you. I hadn't thought about that comment. Certainly adds to a discussion about how to maybe control X algorithmically to avoid crossed spreads. $\endgroup$
    – Attack68
    Jan 29, 2021 at 20:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.